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Steinberg-like theorems for backbone colouring
J. Araujo, Frédéric Havet, M. Schmitt
To cite this version:
J. Araujo, Frédéric Havet, M. Schmitt. Steinberg-like theorems for backbone colouring. Discrete
Applied Mathematics, Elsevier, 2018, �10.1016/j.dam.2017.03.009�. �hal-01796713�
Steinberg-like theorems for backbone colouring
J. Araujo
F. Havet
M. Schmitt
Draft of November 21, 2016
Abstract
A function f : V (G) → {1, . . . , k} is a (proper) k-colouring of G if |f (u) − f (v)| ≥ 1, for every edge uv ∈ E(G). The chromatic number χ(G) is the smallest integer k for which there exists a proper k-colouring of G.
Given a graph G and a subgraph H of G, a circular q-backbone colouring f of (G, H) is a k-colouring of G such that q ≤ |c(u) − c(v)| ≤ k − q, for each edge uv ∈ E(H). The circular q-backbone chromatic number of a graph pair (G, H), denoted CBCq(G, H), is the minimum k such that (G, H)
admits a circular q-backbone k-colouring.
Steinberg conjectured that if G is planar and G contains no cycles on 4 or 5 vertices, then χ(G) ≤ 3. If this conjecture is correct, then one could deduce that CBC2(G, H) ≤ 6, for any H ⊆ G.
In this work, we first show that if G is a planar graph containing no cycle on 4 or 5 vertices and H ⊆ G is a forest, then CBC2(G, H) ≤ 7. Then, we prove that if H ⊆ G is a forest whose connected
components are paths, then CBC2(G, H) ≤ 6.
1
Introduction
In this paper, all graphs are considered to be simple and we use standard terminology as the one in [2]. Let G = (V, E) be a graph, and H = (V, E(H)) be a spanning subgraph of G, called its backbone. A k-colouring of G is a mapping f : V → {1, . . . , k}. Let f be a k-colouring of G. It is a proper colouring if |f (u) − f (v)| ≥ 1 for every edge uv ∈ E(G).. It is a q-backbone colouring for (G, H) if f is a proper colouring of G and |f (u) − f (v)| ≥ q for all edges uv ∈ E(H). The chromatic number χ(G) is the smallest integer k for which there exists a proper k-colouring of G. The q-backbone chromatic number of (G, H), denoted by BBCq(G, H), is the smallest integer k for which there exists a q-backbone k-colouring of
(G, H) [3].
Note that if f is a proper k-colouring of G, then the function g : V → {1, . . . , q · k − q + 1} defined by g(v) = q · f (v) − q + 1 is a q-backbone colouring of (G, H), for any spanning subgraph H of G. Moreover it is well-known that if G = H and f is a proper χ(G)-colouring of G, this q-backbone colouring g of (G, H) is optimal. Therefore, since BBCq(H, H) ≤ BBCq(G, H) ≤ BBCq(G, G), we have
q · χ(H) − q + 1 ≤ BBCq(G, H) ≤ q · χ(G) − q + 1. (1)
One can generalise the notion of backbone colouring by allowing a more complicated structure of the colour space. A natural choice is to impose a circular metric on the colours. We can see Zk1 as a cycle
of length k with vertex set {1, . . . , k} together with the graphical distance | · |k. Then |a − b|k ≥ q if and
only if q ≤ |a − b| ≤ k − q. A circular q-backbone k-colouring of G or q-backbone Zk-colouring of (G, H)
is a mapping f : V (G) → Zk such that c(v) 6= c(u), for each edge uv ∈ E(G), and |c(u) − c(v)|k ≥ q
for each edge uv ∈ E(H). The circular q-backbone chromatic number of a graph pair (G, H), denoted CBCq(G, H), is the minimum k such that (G, H) admits a circular q-backbone k-colouring.
Note that if f is a circular q-backbone k-colouring of (G, H), then f is also a q-backbone k-colouring of (G, H). On the other hand, observe that a q-backbone k-colouring f of (G, H) is a circular q-backbone (k + q − 1)-colouring of (G, H). Hence for every graph pair (G, H), where H is a spanning subgraph of G, we have
BBCq(G, H) ≤ CBCq(G, H) ≤ BBCq(G, H) + q − 1. (2)
1Whenever we refer to Z
Combining Inequalities (1) and (2), we observe that
q · χ(H) ≤ CBCq(G, H) ≤ q · χ(G). (3)
In this paper, we focus on the particular case when G is a planar graph and H is a forest (i.e. an acyclic graph). Inequality (1) and the Four-Colour Theorem [1] imply that for any planar graph G and spanning subgraph H, BBCq(G, H) ≤ 3q + 1. However, for q = 2, Broersma et al. [3] conjectured that
this is not best possible if the backbone is a forest.
Conjecture 1. If G is planar and H is a forest in G, then BBC2(G, H) ≤ 6.
This conjecture would be tight even if H is a Hamilton path as there are examples of planar graph G and Hamilton path H in G, for which BBC2(G, H) = 6. Furthermore, Havet et al. [6] proved that given
a planar graph G and a Hamilton path H in G, it is NP-complete to decide whether BBC2(G, H) ≤ 5.
Campos et al. [4] proved Conjecture 1 when H is a tree of diameter at most 4. For larger values of q, Havet et al. [6] proved the following.
Theorem 2. If G is planar and H is a forest in G, then BBCq(G, H) ≤ q + 6.
For q ≥ 4, they also show that Theorem 2 is best possible. On the other hand, they conjecture that if q = 3, Theorem 2 is not best possible.
Conjecture 3. If G is planar and H is a forest in G, then BBC3(G, H) ≤ 8.
Regarding circular backbone colouring, Havet et al. [6] proved the following. Theorem 4. If G is planar and H is a forest in G, then CBCq(G, H) ≤ 2q + 4.
They also conjectured that this upper bound can be reduced by at least 1. Conjecture 5. If G is planar and H is a forest in G, then CBCq(G, H) ≤ 2q + 3.
Observe that Inequalities (1) and (3) imply that Conjectures 1, 3, and 5 for q ≤ 3, hold if G is has a 3-colouring. There are many sufficient conditions implying that a planar graph has a 3-colouring. For example, the celebrated Gr¨otzsch’s Theorem [5] asserts that planar graphs with girth at least 4 admit a 3-colouring.
One of the most famous conjectures on planar graphs was posed in 1976 by Steinberg (see [7]). Conjecture 6 (Steinberg, 1976). If G is a planar graph with no cycles of length 4 or 5, then χ(G) ≤ 3.
Note that this long standing conjecture could be proved by showing that CBC2(G, G) = 6, for a
planar graph G containing no C4nor C5as a subgraph.
In this paper, we give evidences to the above conjectures. We first settle Conjecture 5 for q = 2 when G is a planar graph without cycles of length 4 or 5.
Theorem 7. If G is a planar graph with no cycles of length 4 or 5 and H is a forest in G, then CBC2(G, H) ≤ 7.
We then improve the upper bound when H is a path forest. A forest is a path forest if all its connected components are paths.
Theorem 8. If G is a planar graph without cycles of length 4 or 5, and H is a path forest in G, then CBC2(G, H) ≤ 6.
Hence when G is a planar graph with no cycles of length 4 or 5 and H is a path forest, Conjecture 1 and Conjecture 5 for q = 2 hold. It also implies that Conjecture 5 for q = 3 holds for such graph pair (G, H) thanks to the following lemma.
Lemma 9. Let k and q be positive integers, G a graph and H a subgraph of G. If CBCq(G, H) ≤ q × k,
Proof. Let c be a q-backbone qk-colouring of (G, H). Define k intervals of colours Ij = {(j − 1)q +
1, . . . , jq}, for every 1 ≤ j ≤ k. We now define a proper q0-backbone q0k-colouring of (G, H) c0 as the following: if c(v) ∈ Ij, then c0(v) = c(v) + (j − 1)(q0− q), for every 1 ≤ j ≤ k and v ∈ V (G). Note
that c0 must be a proper colouring of G since c is a proper colouring of G. Moreover, for an edge uv ∈ E(H), we have that c(u) and c(v) cannot lie on the same interval Ij, for some 1 ≤ j ≤ k. Thus,
k − q0= (k − q) + (q0− q) ≥ |c(u) − c(v)| ≥ q + (q0− q) = q0.
Corollary 10. If G is a planar graph without cycles of length 4 or 5, and H is a path forest in G, then CBC3(G, H) ≤ 9.
The remainder of this paper is devoted to the proofs of Theorems 7 and 8. Both results are proved by supposing that a minimal counter-example with respect to the number of vertices exists. In Section 2, we present general properties of such counter-examples that we use in both proofs. Then, we prove Theorems 7 and 8 in Sections 3 and 4, respectively.
2
Preliminaries
In this section, we first introduce some useful definitions and notations. We then establish properties that we shall use to prove the two above-mentioned theorems. We first prove simple properties of a planar graph without cycles on 4 or 5 vertices, then we establish some properties of minimal counter-examples to Theorems 7 and 8.
2.1
Definitions and notations
Let S ⊆ V (G) be a subset of vertices of G and F ⊆ E(G) be a subset its edges. We denote by G[S] the subgraph of G induced by S, by G\F the graph obtained from G by removing the edges in F from its edge set E(G), by G − S the graph G[V (G) \ S], by (G, H) − S the graph pair (G − S, H − S), by (G, H)[S] the graph pair (G[S], H[S]) and by (G, H)\F the graph pair ((G\F ), (H\F )).
Let (G, H) and (G0, H0) be graph pairs such that H ⊆ G and H0 ⊆ G0. We say that (G0, H0) is a
subpair of (G, H) if H0 ⊆ H and G0 ⊆ G. We say that (G0, H0) is a proper subpair of (G, H) if it is a
subpair of (G, H) and H0⊂ H or G0 ⊂ G. Note that the previous or condition is not exclusive.
A graph pair (G, H) is (k, 2)-minimal if CBC2(G, H) > k, but CBC2(G0, H0) ≤ k for every proper
subpair (G0, H0) of (G, H).
For every colour 1 ≤ a ≤ k, let [a]k be the set of colours b ∈ {1, . . . , k} satisfying |a − b| < 2 or
|a − b| > k − 2. Note that [a]k has always 3 colours.
We say that c is a partial circular 2-backbone k-colouring of (G, H) if c is a circular 2-backbone k-colouring of a subpair of (G, H). Let c be a partial circular 2-backbone k-colouring of a graph pair (G, H). We say that a colour a is available (or possible) at vertex v ∈ V (G) if none of its neighbours in G is coloured a and none of its neighbours in H has a colour in [a]k. We denote by Avc(G, H, v) the set
of available colours at v in c.
We emphasize that this definition of available colour does not require that the vertex v is not coloured. If v is already coloured, observe that it can be recoloured by any available colour and we obtain another feasible partial circular 2-backbone colouring of the same subpair of (G, H).
Similarly, we say that a colour a is forbidden at vertex v due to S ⊆ NG(v) if there exists u ∈ S ∩NG(v)
coloured a, or if there exists u ∈ S ∩ NH(v) with colour in [a]k. When we consider S = NG(v), we simply
mention forbidden at vertex v. We denote the set of forbidden colours at v in c by Fbc(G, H, v).
We also define a colour a is strongly forbidden at vertex v due to S ⊆ NG(v) if one cannot obtain
a partial circular 2-backbone k-colouring by assigning to v the colour a and (re)colouring each vertex u ∈ S with one colour in Avc(G, H, u).
We often omit G, H and c from these notations when they are clear in the context. Observe that Av(v) = Zk\ Fb(v).
In the following sections, the graph H is a forest. Then, a vertex of degree 0 (resp. 1, at least 2) in H is called an isolated vertex (resp. a leaf, a node). We also denote by leaf(H) the number of leaves in H, by isol(H) the number of isolated vertices in H and by comp(H) the number of its connected components.
Planar graphs with no cycles of length 4 or 5.
A well-known result on planar graphs is the Euler’s Formula:
Theorem 11 (Euler’s Formula). If G is a connected plane graph, then |V (G)| − |E(G)| + |F (G)| = 2. Let τ (G) denote the number of triangles of a graph G.
Lemma 12. If G is a planar graph with no C4, then the three statements hold:
(i) every two (not identical) triangles do not share an edge; (ii) every vertex v ∈ V (G) belongs to at most bdG(v)/2c triangles;
(iii) 3τ (G) ≤ |E(G)|.
Proof. To prove (i), observe that two triangles with different vertex sets cannot share at least one edge, otherwise there would be a C4in G. (ii) follows directly from (i) by observing that the graph induced by
the neighbourhood of v contains no path of length 2, since there is no 4-cycles. Finally, we derive (iii) from (ii). Indeed (ii) implies 3τ (G) ≤P
v∈V (G)bdG(v)/2c. SincePv∈V (G)dG(v)/2 = |E(G)|, we obtain
3τ (G) ≤ |E(G)|.
Lemma 13. If G is a connected planar graph with no cycles of length 4 or 5 and G 6= K3, then
|E(G)| ≤ 2|V (G)| − 4.
Proof. Without loss of generality, we may assume tat G is embedded in the plane. By Euler’s Formula, we have 6|E(G)| = 6|V (G)| + 6|F (G)| − 12. Hence
4|E(G)| = 6|V (G)| − X
f ∈F (G)
(d(f ) − 6) − 12 ≤ 6|V (G)| + 3|F3(G)| − 12
where F3(G) is the set of 3-faces in G. Since G 6= K3, we observe that |F3(G)| ≤ τ (G), and then
3|F3(G)| ≤ |E(G)|, by Lemma 12-(iii). Hence, 3|E(G)| ≤ 6|V (G)| − 12, that is |E(G)| ≤ 2|V (G)| − 4.
Properties of (k, 2)-minimal pairs
Lemma 14. Let (G, H) be a (k, 2)-minimal pair and c be a partial 2-backbone k-colouring of (G, H). If uv ∈ E(H) and 1 ≤ | Av(u)| ≤ 3, then at most 4 − | Av(u)| colours are strongly forbidden at v due to u. Proof. Observe that a colour cannot be assigned to v due to u with uv ∈ E(H) if and only if it is in the set F =T
a∈Av(u)[a]k. Note that this set F is maximized when Av(u) has consecutive colours. Thus, F
has at most 3 colours when | Av(u)| = 1, F has at most 2 colours when | Av(u)| = 2 and F has at most 1 colour when | Av(u)| = 3.
The total degree of a vertex v in (G, H) is dt(v, G, H) = dG(v)+2dH(v). We often simplify the notation
dt(v, G, H) to dt(v) when G and H are clear from the context. Note that dt(v) is an upper bound on the maximum number of forbidden colours at v, when extending a circular 2-backbone k-colouring of (G, H) − v to v.
Lemma 15. If (G, H) is a (k, 2)-minimal pair, then G is connected.
Proof. By contradiction, suppose that G is not connected and let C ⊂ V (G) be a connected component of G. Since (G, H) is (k, 2)-minimal, observe that (G, H) − C and (G, H)[C] admit circular 2-backbone k-colourings c and c0, respectively. Combining c and c0, one obtains a circular 2-backbone k-colouring of (G, H). This is a contradiction to the hypothesis that CBC2(G, H) > k.
Lemma 16. If (G, H) is a (k, 2)-minimal pair, then dt(v) ≥ k, for all v ∈ V (G).
Proof. Assume for a contradiction that there is a vertex v such that dt(v) < k. By minimality of (G, H), (G, H) − v admits a circular 2-backbone k-colouring c. Now at most dt(v) colours are forbidden at v. Hence c can be extended into a circular 2-backbone k-colouring of (G, H), a contradiction.
Lemma 17. Let (G, H) be a (k, 2)-minimal pair. If uv ∈ E(H), then dt(u) + dt(v) ≥ 2k + 2.
Proof. Let x ∈ {u, v}. Set r(x) = dt(x) − k. If r(x) ≥ 2, then we have the result by Lemma 16. So we
may assume that r(x) ≤ 1. By minimality of (G, H), there exists a circular 2-backbone k-colouring c of (G, H) − {u, v}. Since uv ∈ E(H), the total degree of x in (G, H) − {u, v} is k + r(x) − 3. Hence there is a set Av(x) of k − (k + r(x) − 3) = 3 − r(x) available colours at x.
Let F be the set of strongly forbidden colors at u due to v. By Lemma 14, |F | ≤ r(u) + 1. Since (G, H) is a (k, 2)-minimal pair, the colouring c cannot be modified into a circular 2-backbone k-colouring of (G, H). Thus |F | ≥ | Av(x)|, so r(u) + 1 ≥ 3 − r(v). Hence dt(u) + dt(v) ≥ 2k + 2.
3
Forest backbone
The aim of this section is to prove the following.
Theorem 7. If G is a planar graph with no cycles of length 4 or 5 and H is a forest in G, then CBC2(G, H) ≤ 7.
From this point to the end of Section 3, let (G, H) be a minimal counter-example to Theorem 8 and whenever we use the notation [a], when a is a colour, we mean [a]7. By Lemmas 15, 16 and 17, observe
that G must be connected, each vertex must have total degree at least 7 and, for each edge uv ∈ H, dt(u) + dt(v) ≥ 16. We now prove extra properties of such a counter-example.
Lemma 18. Let P = uvw be a path in H and c be a circular 2-backbone 7-colouring of (G, H)−{u, v, w}. If | Av(u)| = 3 and | Av(w)| = 3, then at most 3 colours are strongly forbidden at v due to {u, w} (no matter whether uw ∈ E(G)).
Proof. Suppose | Av(u)| = 3 and | Av(w)| = 3. Note that if we show that at most 3 colours are strongly forbidden when uw ∈ E(G), then we also show that the same holds when uw /∈ E(G). In fact, we have the same amount of colours and one constraint less, the one that u and w must receive disjoint colours, to extend c to (G, H). So let us assume uw ∈ E(G).
Let A1= Av(u) and A2= av(w). Let N be the set colours of Z7that are consecutive to some colour
of Av(u). Observe that |N | ≥ 4 and that |N | = 4 if and only if Av(u) = {i, i + 2, i + 4} for some i and N = Z7\ Av(u).
If N ∩ Av(w) = ∅, then |N | = |Z7| − | Av(w)| = 4. It follows that Av(w) = Av(u) = {i, i + 2, i + 4}
for some i, and the set of strongly forbidden colours at v is {i + 1, i + 3} (recall that uw ∈ E(G)). Henceforth we may assume that N and Av(w) intersect. Thus there are two consecutive colours a1∈ Av(u) and a2∈ Av(w). Without loss of generality, we may assume that a2= a1+ 1. Observe that
the three colours not in [a1] ∪ [a2] are not strongly forbidden at v.
If a1− 1 or a2+ 1 is not strongly forbidden, we have the result. So we may assume that both a1− 1
and a2+ 1 are strongly forbidden. Then Av(u) = [a1− 1] or [a1− 1] contains exactly two colours of
each set Av(u) and Av(w) and the colour of Av(u) \ [a1− 1] is the same as the color of Av(w) \ [a1− 1].
Similarly, either Av(w) = [a2+1] or [a2+1] contains exactly two colours of Av(u) and exactly two colours
of Av(w) and the colour of Av(u) \ [a2+ 1] is the same as the colour of Av(w) \ [a2+ 1]. Consequently,
we necessarily have Av(u) = [a1− 1] and Av(w) = [a2+ 1]. Now a1 and a2 are not strongly forbidden
at v and there are just two strongly forbidden colours at v due to {u, w}.
Lemma 19. If S = {s1, . . . , st} is the set of neighbours of v in H such that dt(si) = 7, for every
1 ≤ i ≤ t, then dt(v) ≥ 8 + t.
Proof. The total degree of v in (G, H) is dt(v) = 3t + ˜d with ˜d the total degree of v in (G, H) − S. In
particular, if t ≥ 4, then dt(v) ≥ 3t ≥ 8 + t. Therefore, we may assume that t ≤ 3. If t = 1, then
Lemma 17 yields the results. Henceforth, we now assume t ∈ {2, 3}.
By minimality of (G, H), there exists a circular 2-backbone 7-colouring c of (G, H) − (S ∪ {v}). In (G, H) − (S ∪ {v}), the total degree of si is at most 4. Hence, there is a set Ai of at least 3 available
colours at si.
Let F1,2 (resp. F3) be the set of strongly forbidden colours at v by {s1, s2} (resp. {s3}). By
Lemmas 17 and 18, we have |F3| ≤ 4 − |A3| ≤ 1 and |F1,2| ≤ 3.
Since (G, H) is minimal, then all seven colours must be strongly forbidden at v. If t = 2, this yields 7 ≤ ˜d + |F1,2|. But dt(v) = ˜d + 6. So dt(v) ≥ 13 − |F1,2| ≥ 10.
If t = 3, this yields 7 ≤ ˜d + |F1,2| + |F3|. But dt(v) = ˜d + 9. So dt(v) ≥ 16 − |F1,2| − |F3| ≥ 11.
Corollary 20. If (G, H) be a (7, 2)-minimal pair such that G is a planar graph and H is a spanning forest of G, then
X
v∈V (G)
dt(v) ≥ 8|V (G)| − isol(H)
.
Proof. Recall that Lemma 16 states that dt(v) ≥ 7, for every v ∈ V (G). In particular, if v is an isolated
vertex of H, then dt(v) ≥ 7. If v is not isolated and dt(v) = 7, then v each neighbour u of v in H has
total degree at least 9, by Lemma 17.
Let u be a vertex such that dt(u) ≥ 9 and define t(u) as the number of neighbours v of u in H such
that dt(v) = 7. Lemma 19 states that dt(u) ≥ 8 + t(u). In particular, dt(u) +P
v∈NH(u)d
t(v) ≥ 8.
Thus, let S9+ (resp. S8, S7) be the set of non-isolated vertices with total degree at least 9 (resp.
exactly 8, exactly 7). Let SI denote the set of isolated vertices of H. Observe that {S9+, S8, S7, SI} form
a partition of V (G) and then:
X v∈V (G) dt(v) = X v∈S9+ dt(v) + X v∈S8 dt(v) + X v∈S7 dt(v) + X v∈SI dt(v) ≥ 8 · |S8| + 8 · |S9+∪ S7| + 7 · isol(H) = 8 · |V (G)| − isol(H)
Proposition 21. Let G be a connected plane graph and H be a spanning forest of G. Then X
v∈V (G)
(2dt(v) − 14) + X
f ∈F (G)
(d(f ) − 6) = −12 − 8 comp(H).
Proof. Since H is a spanning forest of G, note that |E(H)| ≤ |V (G)| − comp(H). Consequently, by using Euler’s Formula, we have that:
X v∈V (G) (2dt(v) − 14) + X f ∈F (G) (d(f ) − 6) = 2 X v∈V (G) dG(v) + 4 X v∈V (G) dH(v) − 14|V (G)| + 2|E(G)| − 6|F (G)|
= 4|E(G)| + 8|E(H)| − 14|V (G)| + 2|E(G)| − 6|F (G)|
≤ 6|E(G)| − 6|V (G)| − 6|F (G)| − 8 comp(H) = −12 − 8 comp(H).
We are now ready to prove Theorem 7.
Proof of Theorem 7. Let (G, H) be a minimal counter-example with respect to the number of vertices. It is easy to check that the theorem holds when |V (G)| ≤ 3. Hence we have |V (G)| ≥ 4, and (G, H) is (7, 2)-minimal. By Lemma 15, G must be connected.
Set Σ :=P
v∈V (G)(2d
t(v)−14)+P
f ∈F (G)(d(f )−6). Proposition 21 states that Σ ≤ −12−8 comp(H).
We prove that such a counter-example does not exist by finding a contraction to this fact. By Corollary 20, recall that P
v∈V (G)d
t(v) ≥ 8|V (G)| − isol(H). Consequently, P
v∈V (G)(2d t(v) −
14) ≥ 2|V (G)| − 2 isol(H).
Since G has no cycle of length 4 or 5, P
f ∈F (G)(d(f ) − 6) ≥ −3τ (G) ≥ −|E(G)| ≥ 4 − 2|V (G)|, by
Lemmas 12 and 13. Combining these inequalities, we get Σ ≥ 4 − 2 isol(H) ≥ 4 − 2 comp(H), because isol(H) ≤ comp(H) for any graph H.
4
Path forest backbone
Recall that a path forest is a forest whose connected components are paths. The aim of this section is to prove the following.
Theorem 8. If G is a planar graph without cycles of length 4 and 5, and H is a path forest, then CBC2(G, H) ≤ 6.
In order to proof this theorem, we also consider a minimum counter-example of Theorem 8. We also consider that G is embedded in the plane, so that its face set is defined. In Subsection 4.1, we establish some properties of this counter-example. Then, in Subsection 4.2, we use these properties to derive a contradiction via the Discharging Method.
We first need some definitions. For convenience, in this section we often abbreviate circular 2-backbone 6-colouring by colouring. With a slight abuse of notation, we also refer to the set of colours as Z6 so that the modulo operation is already defined (recall that initially the set of colours of a proper
k-colouring is {1, . . . , k}). We also write [c] instead of [c]6. Moreover, we say that two colours a and b are opposite if |a − b|6= 3.
Recall that (G0, H0) is a subpair of (G, H) if H0⊆ H and G0 ⊆ G. We say that (G0, H0) is an induced subpair of (G, H) if V (H0) = V (G0) and H0 = H[V (G0)] and G0 = G[V (G0)] are the corresponding induced subgraphs. A configuration of (G, H) is an induced subpair in which the total degree of some of the vertices is constrained to given values.
In the remainder of this section, we represent a configuration (G0, H0) by a representation of the graph G0 with the edges of H0 in bold and a number t inside the circle corresponding to vertex v if the total degree of v must be equal to t, except for the last configuration where we use ≤ 8 inside a circle to represent that the corresponding vertex has total degree at most 8.
Let C = (G0, H0) be a configuration of (G, H). A vertex of C whose degree is not constrained to some value is called external. The set of external vertices of C is by Ext(C). The vertices of C whose degree is fixed to a number are the internal vertices of C and they form the set Int(C). Thus, Ext(C) and Int(C) form a partition of V (G0). We emphasize that, in all figures of configurations in this section, we depict all neighbours of an internal vertex.
In order to reach a contradiction and prove that no minimum counter-example to Theorem 8 exists, we show that several configurations are forbidden. By forbidden we mean that for these configurations one may extend some circular 2-backbone 6-colouring of (G, H) − Int(C) to (G, H).
One last notion we require is the following: given a configuration C of (G, H) and a vertex x ∈ Ext(C), we say that two colourings c1, c2of (G, H) − Int(C) are C-twin at x, if c1(x) 6= c2(x) and c1(x0) = c2(x0),
for every x0∈ Ext(C) \ {x} (recall that by colouring we now mean circular 2-backbone 6-colouring).
4.1
Properties of a minimal counter-example
In the remainder of this section, (G, H) is always a minimum counter-example to Theorem 8. It means that G is planar containing no cycles on 4 or 5 vertices, H ⊆ G is a spanning path forest of G and (G, H) is a minimal (6, 2)-pair. In particular, recall that for every vertex set S ⊆ V (G), (G, H) − S admits a circular 2-backbone 6-colouring and (G, H)\e also does, for every edge e ∈ E(G).
We now establish some properties of (G, H). First recall that Lemmas 12, 16 and 17 yield the following.
Property 22. Every vertex v ∈ V (G) is incident to at most bd(v)/2c triangles. Property 23. For each vertex v ∈ V (G), dt(v) ≥ 6.
Property 24. If uv ∈ E(H), then dt(u) + dt(v) ≥ 14.
We now study more deeply the structure of (G, H). Let C0 be a configuration consisting of path
uvw in H such that uw /∈ E(G) and dt(v) = 6, i.e. Int(C) = {v} and Ext(C) = {u, w}. Let C0 0 be the
configuration C0 when the edge uw ∈ E(G) exists (see Figure 1).
Property 25. If (G, H) contains a configuration C ∈ {C0, C00}, then
6 6
Figure 1: The configurations C0 (left) and C00 (right).
(ii) if C = C00 and u is a node in H, then dt(u) ≥ 9.
Proof. (i) Suppose, for a contradiction, that there exist two colourings c and c0 that are C-twin at u. One colour of {c(u), c0(u)}, say c(u), is not opposite to c(w) = c0(w). Thus F = [c(u)] ∪ [c(w)] 6= Z6.
Hence, choosing c(v) in Z6\ F , we obtain a colouring of (G, H), a contradiction.
(ii) Suppose again for a contradiction that C = C00, u is a node and dt(u) ≤ 8. Due to the minimality
of (G, H), (G0, H0) = (G, H) \ {uv, uw} has a circular 2-backbone 6-colouring c. Since dt(G0,H0)(u) ≤ 4,
vertex u has at least two available colours with respect to c. Moreover, since u is a node (i.e. a vertex of degree at least 2 in H) there are three consecutive colours in Z6 that are forbidden to u. So there are
at least two available colours at u that are not opposite. Hence one can assign to u one of its available colours that does not belong to {c(w), c(w) + 3} and there would be at least one available colour to extend c to v. Consequently there is a circular 2-backbone 6-colouring of (G, H), a contradiction.
Let C1 be a configuration on 5 vertices u0, u, v, v0 and u00, such that u0uvv0 form an induced path
P1 in G and in H, u00 is not a neighbour of a vertex of P1 in H, but it has exactly two neighbours of
P1 in G: u and v. Moreover, the total degree of u and v must be equal to 7, i.e. Int(C1) = {u, v} (see
Figure 2).
7 7
Figure 2: The configuration C1.
Property 26. If (G, H) contains a configuration C1, then
(i) there is no pair of C1-twin colourings at u0, and
(ii) dt(u0) ≥ 8.
Proof. (i) Suppose for a contradiction that (G, H) − {u, v} admits two C1-twin colourings c and c0 at u0.
One of them, say c, satisfies c(u0) 6= c(v0). Note that c(u00) is forbidden to u and v. Consequently, with respect to the colouring c, the sets Av(u) and Av(v) of available colours at u and v have size at least 2 and they are not equal, since c(u0) 6= c(v0). Thus we can choose c(u) ∈ Av(u) and c(v) ∈ Av(v) so that |c(u) − c(v)|6≥ 2. We then obtain a colouring of (G, H), a contradiction.
(ii) Suppose for a contradiction that dt (G,H)(u
0) ≤ 7. The graph (G, H) − {u, v} admits a circular
2-backbone 6-colouring c. The number of available colours at u0 in this colouring is at least 2. Hence, we can extend c into two C1-twin colourings of (G, H) − {u, v} at u0. This contradicts (i).
Let C2 be a configuration on eight vertices u0, u, v, v0, w, u00, v00 and z, such that u0uvv0w form
an induced path P2 in G and in H, the vertices u00, v00 and z are not neighbours of a vertex of P2 in
H, but u00 and v00 have exactly two neighbours of P1 in G: u and v, v and v0, respectively; and the
only neighbour of z in P2 is v0. Moreover, tdt(u) = 7 and dt(v) = dt(v0) = 8. The configuration C20 is
obtained from C2 by removing z and changing on the total degree of v0 to 7 (see Figure 3). Note that
Int(C2) = Int(C20) = {u, v, v0}.
8
7 8 7 8 7
Figure 3: The configurations C2 (left) and C20 (right).
(i) there is no pair of C-twin colourings at u0, and
(ii) dt(u0) ≥ 8.
Proof. (i) Suppose for a contradiction that there are two C-twin colourings at u0, say c and c0.
First we claim that we can extend c and c0 to v0 with a colour c(v0) = c0(v0) which is not opposite to c(v00). Indeed, if c(v00) ∈ [c(w)], one we can choose c(v0) = c0(v0) in {c(w) + 2, c(w) + 3, c(w) + 4} \ {c(z), c(v00) + 3}. Otherwise, c(v00) /∈ [c(w)] and in this case one can choose c(v0) = c0(v0) in
{c(w) + 2, c(w) + 3, c(w) + 4} \ {c(v00), c(z)}.
Consider now that c and c0 were extended to v0 so that c(v0) = c0(v0) is not opposite to c(v00). If |c(v0) − c(v00)|
6 = 1, then c(v00) is forbidden to v by both v00 and v0. So this case behaves exactly
as if the edge vv00 did not exist in C and we have a configuration equivalent C1. Thus, similarly to
the proof of Property 26-(i), one can derive that no pair of C-twin colourings at u0 exists. Henceforth |c(v0) − c(v00)|
6 = 2. Without loss of generality, assume that c(v00) = 1 and c(v0) = 3. Observe that
Av(v) = {5, 6} \ {c(u00)} for both colourings c and c0.
We distinguish few cases according to the value of c(u00). In each of them, we extend c to a colouring
of (G, H), which is a contradiction.
• Case 1: c(u00) = 1. Recall that c and c0are C-twin colourings at u0. Thus, without loss of generality,
assume that c(u0) 6= 3. One can then choose c(u) ∈ [3] \ [c(u0)] and c(v) ∈ {5, 6} \ [c(u)]. • Case 2: c(u00
) ∈ {2, 3}. The set Av(u) of available colours at u is Z6\ [c(u0)] ∪ {c(u00)}, in the
colouring c. Observe that Av(v) = {5, 6}, | Av(u)| ≥ 2 and Av(u) 6= Av(v), since u0 forbids three consecutive colours and c(u00) ∈ {2, 3}. Thus one can find c(u) ∈ Av(u) and c(v) ∈ Av(v) so that |c(u) − c(v)|6≥ 2.
• Case 3: c(u00) ∈ {4, 6}. Without loss of generality, we may assume that c(u0) 6= 2. Then, choose
c(u) ∈ [2] \ [c(u0)] and set c(v) = 5.
• Case 4: c(u00) = 5. Without loss of generality, assume that c(u0) 6= 3. Then, one can choose
c(u) ∈ [3] \ [c(u0)] and set c(v) = 6.
(ii) Suppose for a contradiction that dt(G,H)(u0) ≤ 7. The graph (G, H) − {u, v, u0, v0} admits a colouring c. The number of available colours at u0 in c is at least 2. Hence, we can extend c into two C-twin colourings at u0. This contradicts (i).
Let C3be a configuration of (G, H) on four vertices u, w, v and z such that the only edges of H are
uw and vw and the edges in E(G) \ E(H) are uv and wz. Moreover, the total degree of w must be 7 (see Figure 4).
7
Figure 4: Configuration C3
(i) dt(u) ≥ 7, and
(ii) if u is a node, then dt(u) ≥ 8.
Proof. Let c be a colouring of (G0, H0) = (G, H) \ {uw, uv}. Without loss of generality, we may assume that c(v) = 1 and c(z) /∈ {2, 3}.
Suppose for a contradiction that dt
(G,H)(u) ≤ 6 or that u is a node and d t
(G,H)(u) = 7. If d t
(G,H)(u) ≤ 6,
u has at least four available colours in c, otherwise u has three available colours which are consecutive. In both cases, we can assign to u a colour c(u) ∈ {2, 5, 6}. If c(u) ∈ {5, 6}, then setting c(w) = 3, we obtain a colouring of (G, H), a contradiction. If c(u) = 2, then choosing c(w) ∈ {4, 5} \ {c(z)}, we obtain a colouring of (G, H), a contradiction.
Let C4 be a configuration of (G, H) on eight vertices u, u0, u00, v, v0, v00, w and z such that the only
edges of H are u0u, uw, wv and vv0 and the only edges in E(G) \ E(H) are u00u, uv, vv00 and wz. The vertices of Int(C4) are u, w and v and they must have total degree equal to 8, 7 and 8, respectively (see
Figure 5).
7
8 8
Figure 5: Configuration C4
Property 29. If (G, H) contains a configuration C4, then
(i) dt(u0) ≥ 8, and
(ii) if u0u00∈ E(G − H) and u0 is a node, then dt(u0) ≥ 9.
Proof. Let us first prove a claim that we use to prove both statements.
We claim that there exists no colouring c of (G, H) − {u, v, w} such that c(u00) 6= c(u0) + 3, no matter whether u0u00 ∈ E(G) \ E(H). Let us prove this claim by contradiction and let c such a colouring.
Observe that | Av(u)| ≥ 2 and | Av(v)| ≥ 2, and that Av(u) contains two consecutive colours, since c(u0) 6= c(u00) + 3. Without loss of generality, we assume that {1, 2} ⊆ A(u) and that c(z) ∈ {2, 3, 4}. Since | Av(v)| ≥ 2, we first extend the colouring c to v by colouring v with a colour c(v) 6= 5. Then, in order to reach a contradiction, we show that for every possible colour c(v) 6= 5 one can extend c into a colouring of (G, H):
• Case 1: c(v) ∈ {1, 2, 3}. Set c(w) = 5 and choose c(u) ∈ {1, 2} \ {c(v)}. • Case 2: c(v) = 4. Set c(u) = 2 and c(w) = 6.
• Case 3: c(v) = 6. Set c(u) = 1 and choose c(w) ∈ {3, 4} \ {c(z)}. This completes the proof of the claim.
(i) By contradiction, suppose that dt(u0) ≤ 7. Let c be a colouring of (G, H) − {u, v, w}. Since u is not coloured by c and u0u ∈ E(H), vertex u0 has at least two available colours in c. Thus one can, if necessary, recolour u0 so that c(u00) 6= c(u0) + 3, contradicting the previous claim.
(ii) By contradiction, suppose that u0u00 ∈ E(G − H) and u0 is a node, and dt(u0) ≤ 8. Let c be a
colouring of (G, H)−{u, v, w} and t0be the neighbour of u0in H that is distinct from u. By definition, we have that c(t0) /∈ [c(u0)] and by the previous claim, c(u00) = c(u0) + 3. Combining these facts, one deduces
that c(u00) ∈ [c(t0)]. Moreover, since dt(u0) ≤ 8, u0 has at most one neighbour in G − {u, u00, t0} that is
already coloured by c. Therefore, since u is not yet coloured, u0 has at least two available colours in c. Hence, one can recolour u0 with a colour distinct from c(u00) + 3, contradicting to the above claim.
4.2
Proof of Theorem 8
The proof uses the Discharging Method. As again consider in this section that (G, H) is a minimal counter-example to Theorem 8 and we also consider that G is embedded in the plane, so that we can refer to its faces. We first assign a initial weight each vertex and face of G, and prove that the total initial weight is negative. We then apply some discharging rules that do not change the total weight. Finally, using the properties established in the previous section, we prove that the final weight of each vertex and face of G is non-negative, which contradicts the negativity of the total weight. This implies that no such a counter-example exists and proves Theorem 8.
4.2.1 Initial weight
Let us define a function initial weights ϕ : V (G) ∪ F (G) → Z of the vertices and faces of G as follows: • If v is an isolated vertex in H, i.e. dH(v) = 0, then ϕ(v) = 2dt(v) − 6;
• If v is a leaf in H, i.e. dH(v) = 1, then ϕ(v) = 2dt(v) − 10;
• If v is a node in H, i.e. dH(v) = 2, then ϕ(v) = 2dt(v) − 14;
• Every face f ∈ F (G) has weight ϕ(f ) = d(f ) − 6.
Observe that the only faces with negative weight are the 3-faces for which the weight is −3, since G has no C4 nor C5. Let Φ =Pv∈V ϕ(v) +
P
f ∈Fϕ(f ) be the total initial weight. The following lemma
shows that Φ is negative.
Lemma 30. The total initial weight Φ equals −12. Proof. We have Φ =X v∈V (2dt(v) − 14) +X f ∈F (d(f ) − 6) + 4 leaf(H) + 8 isol(H).
We shall use Euler’s Formula, and the three following easy facts, whose proof is left to the reader. X v∈V d(v) = X f ∈F d(f ) = 2|E(G)| (4) X v∈V dt(v) = 2|E(G)| + 4|E(H)| (5)
|E(H)| = |V (G)| − leaf(H)/2 − isol(H) (6) By (4) and (5), we have
Φ = 6|E(G)| + 8|E(H)| − 14|V (G)| − 6|F (G)| + 4 leaf(H) + 8 isol(H)
Then using (6), we get Φ = 6|E(G)| − 6|F (G)| − 6|V (G)|. Finally, by Euler’s Formula, we obtain Φ = −12.
4.2.2 Discharging rules
We now apply a set of discharging rules that we apply to convert ϕ into a final function of weights ϕ0in such a way that no charge is lost, i.e. the total charge should be the same in ϕ0.
• Rule 1: Each vertex sends one unit of charge to each of its incident 3-faces.
• Rule 2: Each q-face, for every q ≥ 6, sharing two consecutive backbone edges with a 3-face sends one unit of charge to the node of total degree 6 this 3-face, if it exists.
• Rules 3: For each of the configurations depicted in Figure 6, vertex s sends one unit of charge to vertex t, if ϕ(s) ≥ 4.
7 (a) Rule 3.1 7 8 8 (b) Rule 3.2 7 (c) Rule 3.3 8 7 (d) Rule 3.4
Figure 6: Rules 3. To send charge s must have initial weight at least 4.
6 (a) Rule 4.1 7 8 8 (b) Rule 4.2 7 7 (c) Rule 4.3 8 7 (d) Rule 4.4 8 7 8 (e) Rule 4.5
Figure 7: Rules 4. The dotted line means that the edge belongs to E(G).
• Rules 4: For each of the configurations depicted in Figure 7, vertex s sends one unit of charge to vertex t.
Lemma 31. Whenever one of the Rules 4 is applied, the vertex s has total degree at least 8.
Proof. The total degree of s is imposed to be 8 in Rules 4.2 and 4.4. It is at least 8 in Rules 4.1, 4.3, and 4.5 by Properties 24, 26, and 27, respectively.
Lemma 32. By Rules 3 and 4, a node sends at most two units of charge and a leaf sends at most one unit of charge.
Proof. In Rules 3 and 4, a vertex s sends charge to a vertex a distance at most 2 in H. Moreover, each time its sends to a vertex t at distance 2 in H, then the common neighbour of s and t in H does not send nor receive any charge by Rules 3 or 4. Therefore, by Rules 3 and 4, a vertex sends at most 1 in each direction along its path in H.
4.2.3 Final weight
We now show that all vertices and all faces have a non-negative final weight. Recall that we denote by ϕ0 the final weight function after all applications of the previously described rules.
Lemma 33. Every face f ∈ F (G) has a non-negative final weight.
Proof. If f is a 3-face, then ϕ(f ) = −3, and f receives 1 for each of its vertices by Rule 1. So ϕ0(f ) = 0. If f is not a 3-face, then f has degree at least 6. So we have ϕ(f ) ≥ 0. However, we have to be sure that Rule 2, the only rule to change the weight of faces, cannot create a face with negative weight. Assume that f shares two backbone edges with k 3-faces Ti, 1 ≤ i ≤ k. Recall that no two triangles
share an edge, by Lemma 12. Thus, one can find a cycle C in G by replacing the two backbone edges in Ti∩ f (and their common endvertex) by the third edge of Ti. The cycles obtained from f by taking
k or k − 1 of such shortcuts are cycles of length d(f ) − k and d(f ) − k + 1, respectively. Since G has no cycles of length 4 and 5, it follows that d(f ) − k ≥ 6. Hence ϕ0(f ) = d(f ) − 6 − k ≥ 0.
Lemma 34. If x is an isolated vertex, then ϕ0(x) ≥ 0.
Proof. Observe that the charge of x is ϕ(x) = 2dt(x) − 6 ≥ dt(x), by Property 23. No isolated vertex is
involved in Rules 2, 3 or 4, so x is concerned only by Rule 1. Since x is adjacent to at most bd(x)/2c 3-faces, by Property 22, we deduce that ϕ0(x) ≥ dt(x) − bd(x)/2c ≥ 0.
Lemma 35. If x is a leaf, then ϕ0(x) ≥ 0.
Proof. Recall that ϕ(x) = 2dt(x) − 10. Vertex x sends at most one unit of charge by Rules 3 and 4, by
Lemma 32. By Property 22, x is incident to at most bd(x)/2c 3-faces. Since bd(x)/2c = bdt(x)/2c − 1, we have ϕ0(x) ≥ 2dt(x) − 10 − (bdt(x)/2c − 1) − 1. Thus, if dt(x) ≥ 7 then ϕ0(x) ≥ 0.
Suppose now that dt(x) < 7, so dt(x) = 6 by Property 23. We have ϕ(x) = 2, so x does not send charge by Rules 3. By Lemma 31, x does not send charge by Rules 4. Moreover x is incident to at most two 3-faces. So we have ϕ0(x) ≥ ϕ(x) − 2 ≥ 0.
For nodes, we distinguish several cases according to their total degree. Lemma 36. Let x be a node. If dt(x) ≥ 9, then ϕ0(x) ≥ 0.
Proof. We have ϕ(x) = 2dt(x) − 14 ≥ dt(x) − 5. Now x gives at most 2 by Rules 3 and 4 by Lemma 32,
and 1 to every 3-face. As x is incident to at most bd(x)/2c = bdt(x)/2c − 2 3-faces (by Property 22), we
have ϕ0(x) ≥ dt(x) − 5 − (bdt(x)/2c − 2) − 2 ≥ ddt(x)/2e − 5 ≥ 0.
Lemma 37. Let x be a node. If dt(x) = 6, then ϕ0(x) ≥ 0.
Proof. x sends no charge by Rules 3 because ϕ(x) = −2, nor by Rules 4 by Lemma 31. Moreover, by Rule 4.1, x receives 2 (1 of each of its neighbours). Now x is adjacent to at most one 3-face by Property 22, because d(x) = 2. If x is incident to no 3-face, then it does not gives anything by Rule 1. So ϕ0(x) = ϕ(x) + 2 = 0. If x is incident to a 3-face, then it gives 1 to this 3-face by Rule 1, but it also receives 1 by Rule 2. Hence ϕ0(x) = ϕ(x) + 2 − 1 + 1 = 0.
Lemma 38. Let x be a node. If dt(x) = 7, then ϕ0(x) ≥ 0.
Proof. x sends no charge by Rules 3 because ϕ(x) = 0, nor by Rules 4 by Lemma 31. Moreover, x is adjacent to at most one 3-face by Property 22, because d(x) = 3.
If x is incident to no 3-face, then it does not send any charge, so ϕ0(v) ≥ ϕ(x) = 0. Assume now that x is incident to a 3-face T . Therefore x sends a charge of 1 by Rule 1.
• If T contains the two backbone edges incident to x, then we are in a Configuration C3. Let us
use the notation of the Figure 4, so our vertex x is w. Observe that dt(u) ≥ 7 and dt(v) ≥ 7 by
Property 24.
– If ϕ(u) < 4 and ϕ(v) < 4, then u and v are nodes. Hence dt(u) ≤ 8 and dt(v) ≤ 8. Consequently, by Property 28, dt(u) = dt(v) = 8. Thus, we are in a Configuration C4 with
x = w. Let us use the notation of Figure 5. If u0 is not adjacent to u00, then x receives 1
from u by Rule 4.2. If u0u00 is an edge, then u0 is either a leaf with total degree at least 7 (by Property 28), or a node with total degree at least 9 (by Property 29). In both cases, ϕ(u0) ≥ 4, so x receives 1 from u0 by Rule 3.2. So ϕ0(x) ≥ ϕ(x) − 1 + 1 ≥ 0.
• If T contains only one of the backbone edges incident to x, then we are in Configuration D depicted Figure 8. Let us use the notation of this figure. By Property 24, both u0 and v have total degree
at least 7.
7
Figure 8: Configuration D
– If ϕ(v) ≥ 4, then x receives 1 from v by Rule 3.3. Hence ϕ0(x) ≥ ϕ(x) − 1 + 1 = 0.
– If ϕ(v) = 2, then v is a node, so dt(v) = 8. Let v0 be the neighbour of v in H that is distinct
from x.
∗ If T is the sole 3-face to which v is incident, then x receives 1 from v by Rule 4.4. Hence ϕ0(x) ≥ ϕ(x) − 1 + 1 = 0.
∗ If v is incident to a 3-face distinct from T , then this 3-face contains v0. If ϕ(v0) ≥ 4,
then x receives 1 from v0 by Rule 3.4. Otherwise v is a node and dt(v0) ≤ 8, so we are in Configuration C2or C20 with x = u, and x receives 1 from u0 by Rule 4.5. In both cases,
ϕ0(x) ≥ ϕ(x) − 1 + 1 = 0.
– If ϕ(v) = 0, the v is a node and dt(v) = 7. So we are in Configuration C
1with x = u. Thus
x receives 1 from u0 by Rule 4.5. Hence ϕ0(x) ≥ ϕ(x) − 1 + 1 = 0.
Lemma 39. Let x be a node. If dt(x) = 8, then ϕ0(x) ≥ 0.
Proof. We have ϕ(x) = 2, so it does not send anything by Rule 3. Vertex x sends at most 2 by Rules 4. By Property 22, x is adjacent to at most two 3-faces, because d(x) = 4.
If x is incident to no 3-faces, it sends a charge of at most 2 in total so ϕ0(x) ≥ ϕ(x) − 2 ≥ 0. If x is incident to two 3-faces, then none of the Rules 4 applies. So x sends a charge of at most 2, and ϕ0(x) ≥ ϕ(x) − 2 ≥ 0.
Assume now that x is incident to exactly one 3-face. If x sends a charge of 0 or 1 by Rules 4, then ϕ0(x) ≥ ϕ(x) − 1 − 1 ≥ 0. To complete the proof we shall now prove that x cannot send a charge of 2 by Rules 4.
Suppose by contradiction that it does. Then x is the vertex s for two configurations C and C0 isomorphic to some depicted Figure 7. Observe that C and C0 cannot be both isomorphic to the configurations depicted in Figures 7(b) or 7(d), because x is incident to one 3-face. Hence one of these two configurations, say C, is isomorphic to either C0, C1, C2, or C20, depicted in Figures 7(a), 7(c) and 7(e).
Let t0 be the vertex of C0 corresponding to t in Figure 7. Observe that t0 is not an external vertex of C,
because G has no cycle of length 4 or 5. Consider a colouring c of (G0, H0) = (G, H) \ (Int(C) ∪ {x, t0}). We have dt
(G0,H0)(t0) ≤ dt(G,H)(t0) − 3 ≤ 4. Hence the set A(t0) of available colours at t0 has size at least
2. Furthermore, dt
(G0,H0)(x) ≤ dt(G,H)(x) − 6 = 2, so the set A(x) of available colours at x has size at
least 4. Hence there exist two distinct colours c1 and c2 of A(x) and two colours c01 and c02 in A(t0) (not
necessarily distinct) such that c1∈ [c/ 01] and c2∈ [c/ 02]. Thus there exists two C-twin colourings at x. This
contradicts Properties 25-(i), 26-(i), or 27-(i).
Lemmas 33, 34, 35,36, 37, 38, and 39 show that all faces and vertices have non-negative final weight. Thus Φ ≥ 0. This contradicts Lemma 30 and completes the proof of the theorem.
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